peter.kourzanov@xxxxxxxxx wrote:
> On Wed, Apr 09, 2008 at 06:34:05PM +0100, Andrew Haley wrote:
>> But it's *not* always easily accessible. In this case,
>>
>> int bar(int p[n])
>
> Well, n must be referring a symbol that is in scope, so
> it must be bound to a value at run-time. Otherwise, if its undefined,
> the sizeof should also be undefined...
>
>> n is somewhere in global scope, maybe in another translation unit,
>> and as I point out below you'd have to copy it at the time the array
>> was created.
>
> What's wrong with taking a snapshot of "n" when the function is
> entered? After all, that's what p[n] specifies in the signature?
I was trying to find out if you were serious.
>> So, you're saying it *should* create a hidden copy of the size
>> parameter for use by sizeof? While this would work, it's not very
>> C.
>
> Hhm, not strictly following C99 standard, sure, but following common
> sense, yes. That's what seems to be happening in this valid C99
> snippet:
>
> int main() {
> int s=10,a[s];
> s=11;
> assert(sizeof a == 40);
> }
Mmm, yes.
> I do understand that our idle rumblings will not change the C99
> standard, but, frankly speaking, its very disappointing having to
> write things such as:
>
> void foo(int s,int a[s])
> {
> /* work-around the C99 always returning
> * sizeof(int*) instead of s*sizeof(int)
> */
> int _a[s];
> s=11;
> for (int i=0; i<sizeof _a/sizeof _a[0]; i++);
> }
>
> This one also generates slightly less optimal code...
>
> Maybe it's a good canditate for a GCC extension?
I'm not sure how. C99 says "6.7.5.3 Function
declarators (including prototypes) ... A declaration of a
parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified
pointer to type’’, where the type qualifiers (if any) are those
specified within the [ and ] of the array type derivation."
Andrew.
